Question Number 158175 by mathocean1 last updated on 31/Oct/21
$$\underset{{x}\rightarrow+\infty\:} {{lim}}\frac{{sinx}+{x}}{\mathrm{3}+\mathrm{2}{sinx}}=? \\ $$
Commented by cortano last updated on 01/Nov/21
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}+{x}}{\mathrm{3}+\mathrm{2sin}\:{x}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{sin}\:{x}}{{x}}+\mathrm{1}}{\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2sin}\:{x}}{{x}}} \\ $$$$=\:\frac{\mathrm{0}+\mathrm{1}}{\mathrm{0}+\mathrm{0}}\:=\:\infty \\ $$
Answered by puissant last updated on 31/Oct/21
$${u}=\frac{\mathrm{1}}{{x}}\:;\:{u}\rightarrow\mathrm{0}\:{quand}\:{x}\rightarrow+\infty \\ $$$$\Rightarrow\:\mathscr{L}\:=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{sinx}+{x}}{\mathrm{3}+\mathrm{2}{sinx}}\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left(\frac{\mathrm{1}}{{u}}\right)+\frac{\mathrm{1}}{{u}}}{\mathrm{3}+\mathrm{2}{sin}\left(\frac{\mathrm{1}}{{u}}\right)} \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{u}}}{\mathrm{3}+\frac{\mathrm{2}}{{u}}}\:\left({en}\:{utilisant}\:{le}\:{devellopement}\right. \\ $$$${limit}\acute {{e}}\:{du}\:{sinus}\:{a}\:{l}'{ordre}\:\mathrm{1}.. \\ $$$$\Rightarrow\:\mathscr{L}\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}}{{u}}}{\frac{\mathrm{3}{u}+\mathrm{2}}{{u}}}\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}}{\mathrm{3}{u}+\mathrm{2}}\:=\:\mathrm{1} \\ $$